Four rods each of length I havebeen hinged to form arhombus. Vertex A

1.	Four rods each of length I havebeen hinged to form arhombus. Vertex A is fixed torigid support, vertex C is beingmoved along the x-axis withconstant velocity v as shown inthe figure. The rate at whichvertex B is approaching the x-axis at the moment therhombus is in the form of asquare isanswer only if u know
Answer» $\large\underline {\underline{\bf \purple{Solution}}}:$ Here , Vertex A is in fixed position and vertex C is moving along in the direction of x - axis Let, The velocity of components of vertex B is VY & Vx Given , the ROD B moving towards x - axis ∴ The angle will be 45° as it is a rhombus ∴ Vcos45° = V $\bf_x$ cos45° + V $\bf_y$ cos45° Substituting cos45° = $\sf\dfrac{1}{\sqrt2}$ • $\sf \dfrac{V}{\sqrt2} = \dfrac{V_x}{\sqrt2} + \dfrac{V_y}{\sqrt2}$ • $\sf \dfrac{V}{\sqrt2} = \bigg[\dfrac{V_x+V_y}{\sqrt2}\bigg]$ …… 1 Also, • $\sf V_xcos45^o - V_ycos45^o = 0$ ∵ As its length is not increasing ∴ $\sf V_x=V_y$ Substituting there values in eq 1 • $\sf \dfrac{V}{\sqrt2} = \dfrac{V_y}{\sqrt2}+\dfrac{V_y}{\sqrt2}$ • $\sf \dfrac{V}{\sqrt2} = \dfrac{V_y+V_y}{\sqrt2}$ • $\sf\dfrac{V}{\cancel{\sqrt2}}=\dfrac{2V_y}{\cancel{\sqrt2}}$ • V = $\sf 2V_y$ • $\bf V_y = \dfrac{V}{2}$ $\therefore\rm \underline{Hence,\; option\;C\;is\;your\;answer.}$